Reputation: 330
Does swift type inference not work with function return types?
protocol Vehicle {
func numberOfWheels() -> Int
}
struct Car: Vehicle {
func numberOfWheels() -> Int {
return 4
}
}
struct Bike: Vehicle {
func numberOfWheels() -> Int {
return 2
}
}
struct Truck: Vehicle {
func numberOfWheels() -> Int {
return 8
}
}
struct VehicleFactory {
static func getVehicle<T: Vehicle>(_ vehicleType: T.Type = T.self) -> T? {
let id = identifier(for: T.self)
switch id {
case "Car":
return Car() as? T
case "Bike":
return Bike() as? T
default:
return nil
}
}
private static func identifier(for type: Any.Type) -> String {
String(describing: type)
}
}
let v: Bike = VehicleFactory.getVehicle() // ERROR HERE: Cannot convert value of type 'T?' to specified type 'Bike'
print(v.numberOfWheels())
I am trying this in playground. Why is there an error in above line?
Shouldnt the compiler infer the type to be Bike
from the let v: Bike
declaration?
Upvotes: 0
Views: 259
Reputation: 236360
Not a direct answer to your question. Vadian has already answered but a few notes on your implementation:
(_ vehicleType: T.Type = T.self)
is pointless. You can just omit it.
Second I would simply add init() to your protocol requirements, get rid of the identifier method, change number of wheels to a computed property:
protocol Vehicle {
init()
var numberOfWheels: Int { get }
}
struct Car: Vehicle {
let numberOfWheels = 4
}
struct Bike: Vehicle {
let numberOfWheels = 2
}
struct Truck: Vehicle {
let numberOfWheels = 8
}
struct VehicleFactory {
static func getVehicle<T: Vehicle>() -> T { .init() }
}
let v: Bike = VehicleFactory.getVehicle()
print(v.numberOfWheels) // "2\n"
Upvotes: 2
Reputation: 285079
The problem is that getVehicle
returns an optional, you have to declare
let v: Bike? = VehicleFactory.getVehicle()
Further you have to unwrap v
in the print
line
Upvotes: 3